Let $h(x)=\begin{cases} \sin(x)&\text{for }x < 0 \\\\ \sqrt{x+\pi}&\text{for }x \geq 0 \end{cases}$ Is $h$ continuous at $x=0$ ? Choose 1 answer: Choose 1 answer: (Choice A) A Yes (Choice B) B No
For $h$ to be continuous at $x=0$, we need $\lim_{x\to 0}h(x)$ and $h(0)$ to exist and be equal. Since $0\geq 0$, the rule that applies to $x=0$ is $\sqrt{x+\pi}$. So $h(0)=\sqrt{0+\pi}=\sqrt{\pi}$. Now let's analyze $\lim_{x\to 0}h(x)$. Finding $\lim_{x\to 0^{ +}}h(x)$ For $x$ -values larger than $0$, the appropriate rule for $h(x)$ is $\sqrt{x+\pi}$. Since $\sqrt{x+\pi}$ is continuous for $x\geq0$, any limit in this interval is equal to the expression's value at that limit: $\begin{aligned} &\phantom{=}\lim_{x\to 0^{ +}}h(x) \\\\ &=\lim_{x\to 0^{ +}}[\sqrt{x+\pi}] \gray{\sqrt{x+\pi}\text{ is the rule for }x>0} \\\\ &=\sqrt{0+\pi} \gray{\sqrt{x+\pi}\text{ is continuous at }x=0} \\\\ &=\sqrt{\pi} \end{aligned}$ Finding $\lim_{x\to 0^{ -}}h(x)$ For $x$ -values smaller than $0$, the appropriate rule for $h(x)$ is $\sin(x)$. Since $\sin(x)$ is continuous for all real numbers, any limit is equal to the expression's value at that limit: $\begin{aligned} &\phantom{=}\lim_{x\to 0^{ -}}h(x) \\\\ &=\lim_{x\to 0^{ -}}[\sin(x)] \gray{\sin(x)\text{ is the rule for }x<0} \\\\ &=\sin(0) \gray{\sin(x)\text{ is continuous at }x=0} \\\\ &=0 \end{aligned}$ Conclusion We found that $\lim_{x\to 0^{ +}}h(x)=\sqrt{\pi}$ and $\lim_{x\to 0^{ -}}h(x)=0$. Since the one-sided limits aren't equal, $\lim_{x\to 0}h(x)$ doesn't exist and $h$ isn't continuous at $x=0$.